There are several questions being asked (and an unexplained reference to a field of definition), but the answer to at least one of them is no: Take $G_1 = G_2 = \mathrm{SL}_3(\mathbb{C})$, with a given minimal = maximal parabolic subgroup involving a single simple root subgroup relative to some choice of positive roots; this parabolic clearly won't lie in two distinct maximal ones. (I'm assuming "minimal" excludes a Borel subgroup and "maximal" means proper, though a reductive group might be just a torus.)
So it's a good idea to separate out the more precise question you have in mind and specify whether a field of definition is really involved for the various groups and subgroups involved.
P.S. I've only pointed to the trivial case where there is just one reductive group. But I think your line of questioning will also run into obstacles in situations where the group and subgroup share a common maximal torus but the subgroup is of "pseudo-Levi" type. An example with both $G_1, G_2$ simple of rank 2 occurs when you consider the reductive subgroup of a group of Lie type $G_2$ (in root system notation!) which involves the long roots and has Lie type $A_2$. These pseudo-Levi subgroups are described in terms of proper subsets of vertices in the extended Dynkin diagram which fail to define Levi subgroups of parabolics.
To summarize, there are pairs $G_1 \subset G_2$ and respective parabolic subgroups (all defined over a given subfield of $\mathbb{C}$ such as $\mathbb{Q}$) for which the answers to the various questions asked can be either yes or no. Another example: Take a reductive subgroup $G_1 \cong \mathrm{SL}_2(\mathbb{C})$ of $G_2 = \mathrm{SL}_3(\mathbb{C})$ with a Borel subgroup $P_1$ (involving one simple root $\alpha$) as minimal = maximal parabolic of $G_1$; then $P_1$ liesand the opposite Borel both lie in distinctthe standard maximal parabolicsparabolic $P_2$ of $G_2$ determined by distinct simple rootshaving $-\alpha$ as a root.